Another arithmetic progression problem

[Government$]

Homework Statement

Sum of first three members of increasing arithmetic progression is 30 and sum of their squares is 692. What is the sum of the first 15 members?

The Attempt at a Solution

So i have system of equations:
a1 + a2 + a3 = 30
(a1)^2 + (a2^2) + (a3^2) = 692
----------------------------------
3a1 + 3d = 30
3(a1)^2 +6(a2)d + d^2= 691
---------------------
d = 10 - a1

and if i plug that in 3(a1)^2 +6(a2)d + d^2= 691 i get

2(a1)^2 - 40a1 -192 = 0

now when i find roots of this equation i get 24 and -4 none of them are correct since it says that at the end of the book that a1 = 10 and d=14. Now i do get d=14 with a1=-4 d=-14 with a=-24 but since it is increasing progression i can rule out d=-14, but apparently my solution is not correct. And my solutions satisfy equation 3a1 + 3d = 30 while theirs doesnt.

P.S. This is how they solved this problem:

let a2=a then we get
3a=30 and 3a^2 + 2d^2=692 from that we get a1=10 and d=14 so S15 = 1410

thank you

 

[HallsofIvy]

Homework Statement

Sum of first three members of increasing arithmetic progression is 30 and sum of their squares is 692. What is the sum of the first 15 members?

The Attempt at a Solution

So i have system of equations:
a1 + a2 + a3 = 30
(a1)^2 + (a2^2) + (a3^2) = 692
----------------------------------

Neither of these equations uses the fact that this is an arithmetic progression

3a1 + 3d = 30

I think you are trying to use it here but are doing it incorrectly. Starting from a1 the numbers are a1, a1+ d and a1+ 2d which add to 3a1+ 2d, not "3d".
3a1+ 2d= 30

3(a1)^2 +6(a2)d + d^2= 691

And for the squares, you want a1^2+ (a1+ d)^2+ (a1+ 2d)^2= a1^2+ a1^2+ 4a1d+ d^2+ a1^2+ 4a1d+ 4d^2 which gives
3(a1)^2+ 8a1d+ 4d^2= 692.

---------------------
d = 10 - a1

No, as above, d= 15- (3/2)a1.

and if i plug that in 3(a1)^2 +6(a2)d + d^2= 691 i get

2(a1)^2 - 40a1 -192 = 0

now when i find roots of this equation i get 24 and -4 none of them are correct since it says that at the end of the book that a1 = 10 and d=14. Now i do get d=14 with a1=-4 d=-14 with a=-24 but since it is increasing progression i can rule out d=-14, but apparently my solution is not correct. And my solutions satisfy equation 3a1 + 3d = 30 while theirs doesnt.

P.S. This is how they solved this problem:

let a2=a then we get
3a=30 and 3a^2 + 2d^2=692 from that we get a1=10 and d=14 so S15 = 1410

thank you

 

[Government$]

I don't understand, how come a1 + a2 + a3= a1 + a1 +d + a1 + 2d = 3a1 + 2d and not
3a1 + 3d. What happened to that one d

 

[Curious3141]

I don't understand, how come a1 + a2 + a3= a1 + a1 +d + a1 + 2d = 3a1 + 2d and not
3a1 + 3d. What happened to that one d

3a1 + 3d is right.

 

[Curious3141]

Can we drop the "a1"? I'm used to calling the first term of an AP simply "a".

I get a = -4, d = 14 to satisfy the condition that this is an increasing progression.

The sum of the first 15 terms is 1410.

I'm not sure why the textbook is saying the first term is 10, because that's wrong. But the sum IS 1410.

How did you calculate S15? What answer did you get?

EDIT: The textbook method MASSIVELY simplifies the algebra, so I like it. But remember that what they're calling a is the *second* term, and that's calculated to be 10. So the first term is still 10 - 14 = -4.

The AP goes: -4, 10, 24,...,192 up to the first 15 terms.